Metamath Proof Explorer


Theorem psseq2d

Description: An equality deduction for the proper subclass relationship. (Contributed by NM, 9-Jun-2004)

Ref Expression
Hypothesis psseq1d.1 φA=B
Assertion psseq2d φCACB

Proof

Step Hyp Ref Expression
1 psseq1d.1 φA=B
2 psseq2 A=BCACB
3 1 2 syl φCACB